Short on time? This page is the recap. For the derivation from the limit definition, the graph and worked examples, use the full study guide. This topic is shared by Calculus AB and Calculus BC.
Notation: f′(x), dy/dx and d/dx[…] all mean “the derivative with respect to x”. Fractional exponents are written in brackets, e.g. x^(2/3).
Recap
- The derivative is defined as f′(x) = lim (h → 0) [f(x + h) − f(x)]/h. For x², x³, 1/x and √x this limit gives 2x, 3x², −1/x² and 1/(2√x).
- Those results share one pattern: the power rule.
- The rule works for any constant exponent: whole, negative, fractional or irrational.
- It applies only to a single power of x. Rewrite roots, reciprocals and simple products or quotients of powers first.
- Where xʳ⁻¹ is undefined (often x = 0), the derivative does not exist. Expect a vertical tangent, a cusp or a domain boundary.
Key relationships
| Function | Rewrite | Derivative |
|---|---|---|
| xⁿ (n a positive integer) | — | n·xⁿ⁻¹ |
| x | x¹ | 1 |
| 1/xⁿ | x⁻ⁿ | −n·x⁻ⁿ⁻¹ = −n/xⁿ⁺¹ |
| √x | x^(1/2) | 1/(2√x), x > 0 |
| nth root of xᵐ | x^(m/n) | (m/n)·x^(m/n − 1) |
| xᵃ·xᵇ or xᵃ/xᵇ | xᵃ⁺ᵇ or xᵃ⁻ᵇ | combine first, then use the rule |
Assumptions behind the rule
- The exponent r is a constant. If the variable is in the exponent (as in 2ˣ), this is not a power function.
- The base is x itself. Expressions like (2x + 1)³ need expanding now, or the chain rule later (Unit 3).
- The result holds only where x is in the domain of both xʳ and xʳ⁻¹.
Mistakes to avoid
- −3 − 1 = −2. Wrong: it is −4. Lowering a negative exponent makes it more negative.
- d/dx[1/x³] = 1/(3x²). Rewrite as x⁻³: the answer is −3/x⁴.
- Keeping the old exponent, e.g. (1/2)x^(1/2) for the derivative of √x.
- Differentiating factors separately, e.g. treating x²√x as 2x times 1/(2√x).
- Using the rule on 2ˣ or on a constant like π².
- Missing an undefined derivative at x = 0 for exponents such as 1/3 or 2/3.
Quick self-check
- Find d/dx[x⁻³]. (−3x⁻⁴ = −3/x⁴)
- Let f(x) = x^(5/3). Find f′(8). (f′(x) = (5/3)x^(2/3), so f′(8) = (5/3)(4) = 20/3)
- Find the derivative of √x / x and evaluate it at x = 4. (√x/x = x^(−1/2); derivative −(1/2)x^(−3/2); at 4 this is −(1/2)(1/8) = −1/16)
- Is x^(1/3) differentiable at 0? (No. f′(x) = 1/(3x^(2/3)) has a zero denominator there: vertical tangent.)
Next: practice questions.